A Coil Turned Over

Q. A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The horizontal component of the earth's magnetic field at the place is 3.0 × 10⁻⁵ T. What is the magnitude of the emf induced in the coil, in volts? expects 0.0038 V

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Q. A circular coil of radius 10 cm $A = 0.0314159$ m², 500 $N = 500$ turns and resistance 2 Ω $R = 2$ Ω is placed with its plane perpendicular to the horizontal component $θ_{0} = 0 °$ of the earth's magnetic field. It is rotated about its vertical diameter through 180° $θ_{1} = 180 °$ , $B_{1} = B_{0}$ in 0.25 s $Δ t = 0.25$ s. The horizontal component of the earth's magnetic field at the place is 3.0 × 10⁻⁵ T $B_{0} = 3.0 \times 10^{−5}$ T. What is the magnitude of the emf induced in the coil, in volts?

The playground

Inputs

turns

N

500

resistance

R

2 Ω

time

Δ t

0.25 s

field

B_{0}

3e-05 T

Outputs

emf

ε

The solution

Solved with FaradaysLaw

Given

$N = 500$
$R = 2$ Ω
$Δ t = 0.25$ s
$B_{0} = 3.0 \times 10^{−5}$ T
$A = 0.0314159$ m² from “A circular coil of radius 10 cm”
$θ_{0} = 0 °$ from “plane perpendicular to the horizontal component”
$θ_{1} = 180 °$ from “through 180°”
$B_{1} = B_{0}$ from “through 180°”

Find $ε$ .

Working

flux through a loop(coil, turning, 1) $Φ_{0} = B_{0} \cdot A \cdot \cos θ_{0}$
$Φ_{0} = 3 \times 10^{−5} \cdot 0.0314159 \cdot \cos 0 °$
$Φ_{0} = 9.42477 \times 10^{−7}$ Wb
through 180° $B_{1} = B_{0}$
$B_{1} = 3 \times 10^{−5}$
$B_{1} = 3 \times 10^{−5}$ T
flux through a loop(coil, turning, 2) $Φ_{1} = B_{1} \cdot A \cdot \cos θ_{1}$
$Φ_{1} = 3 \times 10^{−5} \cdot 0.0314159 \cdot \cos 180 °$
$Φ_{1} = −9.42477 \times 10^{−7}$ Wb
emf from the change of flux(coil, turning) $ε = \frac{N \cdot (Φ_{0} - Φ_{1})}{Δ t}$
$ε = \frac{500 \cdot (9.42477 \times 10^{−7} - (−9.42477 \times 10^{−7}))}{0.25}$
$ε = 0.00377$ V
$ε \approx 0.0038$ Vto 2 significant figures

Conclusion

Turning the coil over reverses the flux through it, so the change is twice the flux, shared over every turn of the coil.

The Ruby

1require_relative "../physics"23question :a_coil_turned_over do4  kind :numerical5  describes "A circular coil of radius 10 cm, %{turns} turns and resistance %{resistance} is placed with its plane " \6            "perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its " \7            "vertical diameter through 180° in %{time}. The horizontal component of the earth's magnetic field at " \8            "the place is %{field}. %{emf}, in volts?"9  given turns: "500"10  given resistance: "2 Ω"11  given time: "0.25 s"12  given field: "3.0e-5 T"13  asks emf: "What is the magnitude of the emf induced in the coil"14end

1require_relative "../questions/a_coil_turned_over"2require_relative "../science/electromagnetic_induction/faradays_law"34solve :a_coil_turned_over, using: FaradaysLaw do5  objects :coil6  coil.goes(:turning)78  take turns: coil.turns9  take resistance: coil.resistance10  take time: turning.time_taken11  take field: coil.magnetic_field(at: turning.start)1213  deduce_from "A circular coil of radius 10 cm", coil.area => "0.0314159 m2"14  deduce_from "plane perpendicular to the horizontal component", coil.angle_to_the_field(at: turning.start) => "0 deg"15  deduce_from "through 180°", coil.angle_to_the_field(at: turning.end) => "180 deg",16                              coil.magnetic_field(at: turning.end) => coil.magnetic_field(at: turning.start)1718  emf = coil.induced_emf(during: turning)19  answer emf:, in: "V", figures: 22021  conclusion "Turning the coil over reverses the flux through it, so the change is twice the flux, shared " \22             "over every turn of the coil."23end

1require_relative "../solutions/a_coil_turned_over"23play_with :a_coil_turned_over do4  title "A Coil Turned Over"5end

The laws

quantity	written	units	within	from
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
turns	$N$	—	—	FaradaysLaw
area	$A$	m²	—	FaradaysLaw
resistance	$R$	Ω	0 … ∞	FaradaysLaw
magnetic field	$B$	T	—	FaradaysLaw
angle to the field	$θ$	—	0 … 3.142	FaradaysLaw
magnetic flux	$Φ$	Wb	—	FaradaysLaw
induced emf	$ε$	V	—	FaradaysLaw
induced current	$I$	A	—	FaradaysLaw
displacement of coil during turning	$Δ x$	m	—
position of coil at end of turning	$x_{1}$	m	—
position of coil at start of turning	$x_{0}$	m	—
average velocity of coil during turning	$v_{av}$	m/s	—
time taken of turning	$Δ t$	s	0 … 1e+05
average speed of coil during turning	$c_{av}$	m/s	—
path length of coil during turning	$d$	m	—
magnetic flux of coil at start of turning	$Φ_{0}$	Wb	—
magnetic field of coil at start of turning	$B_{0}$	T	—
area of coil	$A$	m²	—
angle to the field of coil at start of turning	$θ_{0}$	—	0 … 3.142
magnetic flux of coil at end of turning	$Φ_{1}$	Wb	—
magnetic field of coil at end of turning	$B_{1}$	T	—
angle to the field of coil at end of turning	$θ_{1}$	—	0 … 3.142
induced emf of coil during turning	$ε$	V	—
turns of coil	$N$	—	—
induced current of coil during turning	$I$	A	—
resistance of coil	$R$	Ω	0 … ∞

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

Δ x = x_{1} - x_{0}

change of position(turning)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(turning)

c_{av} = \frac{d}{Δ t}

path a time(turning)

Φ_{0} = B_{0} \cdot A \cdot \cos θ_{0}

flux through a loop(coil, turning, 1)

Φ_{1} = B_{1} \cdot A \cdot \cos θ_{1}

flux through a loop(coil, turning, 2)

ε = \frac{N \cdot (Φ_{0} - Φ_{1})}{Δ t}

emf from the change of flux(coil, turning)

I = \frac{ε}{R}

current driven by the emf(coil, turning)

B_{1} = B_{0}

through 180°